Question

2017级专科学校高等数学习题解答

2017级专科学校高等数学习题解答

Answer 1

1.原式=∫(-2,2)f(x-1)d(x-1)
=∫(-3,1)f(x)dx (用x代换x-1)
=∫(-3,0)f(x)dx+∫(0,1)f(x)dx (分解成两个积分的和)
=∫(-3,0)xe^(-x)dx+∫(0,1)√(2x-x^2)dx (利用已知条件)
=[-xe^(-x)]|(-3,0)+∫(-3,0)e^(-x)dx+∫(0,1)√[1-(1-x)²]dx
(第一个积分应用分部积分法)
=-3e³+[-e^(-x)]|(-3,0)+∫(π/2,0)(-cos²t)dt
(在第二个积分中,令1-x=sint)
=-3e³-1+e³+1/2∫(0,π/2)[1+cos(2t)]dt
=-2e³-1+1/2[t+sin(t/2)/2]|(0,π/2)
=-2e³-1+1/2(π/2-0)
=π/4-2e³-1;
2.(1)∵S1=∫(0,a)(ax-x²)dx=(ax²/2-x³/3)|(0,a)=a³/2-a³/3=a³/6
S2=∫(a,1)(x²-ax)dx=(x³/3-ax²/2)|(a,1)=1/3-a/2-a³/3+a³/2=a³/6-a/2+1/3
∴S1+S2=a³/6+a³/6-a/2+1/3=a³/3-a/2+1/3
∵令(S1+S2)'=a²-1/2=0,得a=√2/2 (∵0